Rotational motion formula sheet — torque, MOI, angular momentum quick reference

Question

How do rotational motion quantities relate to their translational counterparts? Provide a complete formula sheet for torque, moment of inertia, and angular momentum problems.

(JEE Main / NEET — Rotational Mechanics quick reference)

Translation-Rotation Analogy Map

flowchart LR
    A["Mass (m)"] -->|Analogy| B["Moment of Inertia (I)"]
    C["Force (F)"] -->|Analogy| D["Torque (tau = r x F)"]
    E["Velocity (v)"] -->|Analogy| F["Angular Velocity (omega)"]
    G["Acceleration (a)"] -->|Analogy| H["Angular Acceleration (alpha)"]
    I["Momentum (p = mv)"] -->|Analogy| J["Angular Momentum (L = I omega)"]
    K["F = ma"] -->|Analogy| L["tau = I alpha"]
    M["KE = mv²/2"] -->|Analogy| N["KE = I omega²/2"]

Solution — Step by Step

Translation	Symbol	Rotation	Symbol
Displacement	$s$	Angular displacement	$\theta$
Velocity	$v$	Angular velocity	$\omega$
Acceleration	$a$	Angular acceleration	$\alpha$
Mass	$m$	Moment of inertia	$I$
Force	$F$	Torque	$\tau$
Momentum	$p = mv$	Angular momentum	$L = I\omega$
Newton’s 2nd law	$F = ma$	$\tau = I\alpha$
Kinetic energy	$\frac{1}{2}mv^2$	$\frac{1}{2}I\omega^2$
Work	$W = Fs$	$W = \tau\theta$
Power	$P = Fv$	$P = \tau\omega$

Every translational formula has a rotational counterpart — replace $m$ with $I$ , $v$ with $\omega$ , $F$ with $\tau$ , and the formula structure stays identical.

Body	Axis	$I$	$k^2/R^2$
Ring	Centre, perpendicular	$MR^2$	1
Disc	Centre, perpendicular	$\frac{1}{2}MR^2$	1/2
Solid sphere	Diameter	$\frac{2}{5}MR^2$	2/5
Hollow sphere	Diameter	$\frac{2}{3}MR^2$	2/3
Solid cylinder	Own axis	$\frac{1}{2}MR^2$	1/2
Thin rod	Centre, perpendicular	$\frac{1}{12}ML^2$	—
Thin rod	End, perpendicular	$\frac{1}{3}ML^2$	—

Pure rolling condition: $v_{CM} = R\omega$

Total KE in rolling: $KE = \frac{1}{2}Mv_{CM}^2\left(1 + \frac{k^2}{R^2}\right)$

Acceleration on incline: $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$

Time to reach bottom of incline: lower $k^2/R^2$ means faster. Order: solid sphere (2/5) beats solid cylinder (1/2) beats hollow sphere (2/3) beats ring (1).

Angular momentum conservation: When $\tau_{net} = 0$ , $L = I\omega = \text{constant}$ .

Why This Works

Rotational motion is not a new set of physics — it is translational motion repackaged around a pivot. Once you see the analogy, you only need to memorise one set of formulas. The MOI ( $I$ ) replaces mass and accounts for how mass is distributed relative to the axis. A ring (all mass at radius $R$ ) has $I = MR^2$ ; a solid sphere (mass spread through volume) has $I = \frac{2}{5}MR^2$ — less rotational inertia per unit mass.

Alternative Method — Dimensional Analysis Check

If you forget a formula during the exam, use dimensional analysis:

Torque = force $\times$ distance = N $\cdot$ m = kg $\cdot$ m $^2$ /s $^2$
MOI = mass $\times$ distance $^2$ = kg $\cdot$ m $^2$
Angular momentum = MOI $\times$ angular velocity = kg $\cdot$ m $^2$ /s

Check: $\tau = I\alpha$ → (kg $\cdot$ m $^2$ /s $^2$ ) = (kg $\cdot$ m $^2$ )(1/s $^2$ ). Dimensions match.

The $k^2/R^2$ ratio is the single most useful number in rolling problems. Memorise these four: ring = 1, disc/cylinder = 1/2, hollow sphere = 2/3, solid sphere = 2/5. With these, you can solve any “which body reaches first” or “KE split” problem in under 30 seconds.

Common Mistake

When applying $\tau = I\alpha$ , always compute $I$ about the actual axis of rotation, not the centre of mass (unless they are the same). For a rod pivoted at one end, $I = ML^2/3$ (not $ML^2/12$ ). Using the wrong axis gives wrong angular acceleration — and everything after that cascades into error.